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4 years ago

What would happen if a bird was a vulture riding on a zebra's back? How would it affect the relationship?

1 answer:

6 0

The zebra could be "hen pecked" if the vulture was a hen bird. On the other hand, the two might get along ok. In UK, it they 'd make a useful combination helping each other to cross increasingly dangerous roads (joke).

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Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football

faltersainse [42]

Answer:

$a=159.32\ m/s^2$

Explanation:

It is given that,

Angular speed of the football spiral, $\omega=6.9\ rev/s=43.35\ rad/s$

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

$v=r\omega$

$v=0.085\times 43.35$

v = 3.68 m/s

The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(3.68)^2}{0.085}$

$a=159.32\ m/s^2$

So, the centripetal acceleration of the laces on the football is $159.32\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

Two 5.0-cm-diameter conducting spheres are 8.0 m apart, and each carries 0.12 mC. Determine (a) the potential on each sphere, (b

blagie [28]

Answer:

Explanation:

Two spheres 10m apart

Each charge on the sphere is 0.12mC= 0.12×10^-3C

Given that the diameter is

5cm=0.05m

Then, the radius is diameter / 2

r=d/2

r=0.05/2

r=0.025m

Potential is given as

V=kq/r

k=9×10^9Nm2/C2

a. Potential on each sphere surface.

They are going to have the same value since the sphere are identical

At the surface of the sphere,

r= 0.025m

V=kq/r

V=9E9×0.12E-3/0.025

V=4.32E7Volts

V=4.32×10^7Volts

b. The electric field at the surface of each sphere will be the same since the charge are identical,

So, Electric field is given as

E=kq/r^2

At the surface

E=9E9×0.12E-3/0.025^2

E=1.728×10^8N/C

c. The potential mid way between the two sphere

The potential difference due to the first sphere

The sphere are 10m apart then the distance mid way is 5m

Then, the radius of the sphere is 0.025m

Total distance from the center is 5.025m

Then,

V=kq/r

V1=9E9×0.12E-3/5.025

V1=2.149×10^5 Volts

Potential difference due to the second sphere is the same as the first, since both are identical

Total potential is V1 +V2

V=2.149E5+2.149E5

V=4.3E5Volts

Total potential at the middle due to the two sphere is 4.3×10^5Volts

d. The potential difference between the sphere at any point is equal to the potential difference found at c

Therefore the potential difference is

V=4.3×10^5Volts

4 0

3 years ago

Consider a motor that exerts a constant torque of 25.0 N⋅m to a horizontal platform whose moment of inertia is 50.0 kg⋅m2 . Assu

Step2247 [10]

To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

$W=\tau \Delta\theta$

In the case of rotational kinetic energy we know that

$KE = \frac{1}{2}I\omega^2$

PART A) $\theta$ is given in revolutions and needs to be in radians therefore

$\theta = 12rev(\frac{2\pi rad}{1rev})$

$\theta = 24\pi rad$

Replacing in the work equation we have to

$W=\tau \Delta\theta$

$W= (25)(24\pi)$

$W = 1884.95J$

PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

$\tau = I \alpha$